Question

If the function $f\left(x\right)=ax+b$ has its own inverse then the ordered pair (a, b) can be.

• A. $(1,0)$
• B. $(-1,0)$
• C. $(-1,1)$
• D. $(1,1)$

Answer 1

The correct option is A $(1,0)$

We are given that,

$f\left(x\right)=f^{-1}\left(x\right)$ or $fo{f}^{-1}\left(x\right)=x.$

$\therefore$ now, For $f^{-1}\left(x\right)$

$f\left(x\right)=y=ax+b.$

$\therefore x=\frac{y-b}{a}$

$\therefore f^2\left(x\right)=\frac{x-b}{a}$

now, $f\left(x\right)=f^{-1}\left(2\right)\Rightarrow ax+b=\frac{x-b}{a}$

$\therefore a^{2}x+ab=x-b$

$\therefore a^{2}x-+ab+b=0$

$\therefore (a^2-1)x+b(x+1)=0$

$x=\frac{-ba+1}{(a^2-1)}$

$\therefore =\frac{-b(a+1)}{(a+1)(a-1)}=\frac{-b}{a-1}=\frac{b}{1-a}$

$\therefore (1-a)x=b$

For $\left(A\right)a=1\Rightarrow b=0$

$\left(B\right)a=-1\Rightarrow b=2x$

$\left(C\right)a=-1\Rightarrow b=2x$

$\left(D\right)a=1\Rightarrow b=0$.

So, the right answer is $(a,b)=(1,0)$